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2022 AIME II Problems/Problem 14

Revision as of 17:06, 3 July 2023 by Bobjoebilly (talk | contribs) (→‎Solution)

Problem

For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$.

Solution

Notice that we must have $a = 1$, otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, every value up to $1000$ can be represented. Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$, and $b-1$ stamps of value $1$, all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.

So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$.

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$. We can get the answer by solving this equation.

$c > \lfloor \frac{c-1}{b} \rfloor + b-1$

$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

$c^2 - 97c + 999 > 0$, $c > 85.3$ or $c < 11.7$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$

$97 > \frac{999}{c}$, $c>10.3$ 

$\text{Case } 1:$ For $10.3 < c < 11.7$, $c = 11$, $\lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97$
$\lfloor \frac{10}{b} \rfloor + b = 8$, $b=7$

$\text{Case } 2: c>85.3$

$c = 86 \rightarrow$ $\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97$
$\lfloor \frac{85}{b} \rfloor + b = 87$, $b=87 > c$, no solution

$c = 87 \rightarrow$ $\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97$
$\lfloor \frac{86}{b} \rfloor + b = 87$, $b=86$ or $1$. We cannot have $b=1$ since it doesn't satisfy $a<b$, and note that if 
$b=86$ we can have $10$ coins of value $c$, $1$ of $b$, and $85$ of $a$ for a total of $96$ coins and still be able to make 
every value from $1$ to $1000$. Thus $c=87$ yields no solution.

$c = 88 \rightarrow$ $\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97$
$\lfloor \frac{87}{b} \rfloor + b = 87$, $b=86$

$c = 89 \rightarrow$ $\lfloor \frac{999}{89} \rfloor + \lfloor \frac{88}{b} \rfloor + b-1 = 97$
$\lfloor \frac{88}{b} \rfloor + b = 87$, $b=86$

The $3$ least values of $c$ are $11$, $88$, $89$. $11 + 88+ 89 = \boxed{\textbf{188}}$

~isabelchen ~edited by bobjoebilly

Video Solution

https://youtu.be/jptMMVCuj34

~MathProblemSolvingSkills.com


See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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